If x and y are non-negative integers, is xy?

Question

If x and y are non-negative integers, is x>y?

(1)  5^x*3^y=9

(2)  \frac{3^x}{4^y}>\frac{4^y}{3^y}

DavidTutorexamPAL · Answer

As all we're given are equations, we'll look for a simplification.
This is a Precise approach.

(1) Since 9 = 3^2 = (3^2)(5^0), then x = 0 and y =2.
Sufficient.

(2) Multiplying the common denominator gives (3^x)(3^y)>(4^y)(4^y) which simplifies to 3^(x+y) > 4^(2y). Since 3 is smaller than 4, then for the inequality to be true, the exponent of 3 must be larger. In other words, x+y > 2y and therefore x > y.
Sufficient.

(D) is our answer

GMATRockstar · Answer

A few good takeaways from this question:

-we can always rewrite integers as their factors, so we can say 9 = 3^2
-we can multiply across inequalities freely (without flipping the sign) when we know the variables are positive
-if a smaller integer is LARGER than a bigger integer and they both have unknown exponents, then the exponent of the smaller integer is obviously "making up" for the difference in value and must be larger than the bigger integer's exponent

The answer is (D).

The flaw with the question : The statements are incompatible. It cannot be that x = 0 and y = 2, and Statement (2) is true. Because if we plug in x = 0 and y = 2, then Statement (2) reads:

1/16 > 16/9
1/16 is not larger than 16/9

So, while there's good takeaways here, the fact that the statements contradict one another do not make this a great GMAT question.

Vaibhavrahul · Answer

If X= 10 & Y= 9 , how it is possible??

Posted from my mobile device

Vaibhavrahul · Answer

Answer will be A
Because the 2nd condition is not sufficient.The condition of X+Y >2Y will not hold always.Thus 2nd condition is not sufficient

chetan2u · Answer

x and y are non-negative integers, so x and y are  \geq{0}

(1)  5^x*3^y=9 
 5^x*3^y=9=3^2 , so x=0 and y=2..... Is x>y?  ---- NO

(2)  \frac{3^x}{4^y}>\frac{4^y}{3^y} 
Since all terms will be positive with whatever values of x and y....
 3^x*3^y>4^y*4^y........3^{(x+y)}>4^{2y} ......
Now  4^{2y}\geq 3^{2y}  .....Equal, when y=0 otherwise '>'
 3^{(x+y)}>4^{2y}\geq 3^{2y} 
 3^{(x+y)}> 3^{2y} 
Equating the powers 
 x+y>2y...x>y ...YES

D, but both statements contradict.

If x and y are non-negative integers, is x>y?

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